If each diode in the figure has a forward bias resistance of $25\,\Omega$ and infinite resistance in reverse bias,what will be the values of the currents $I_1, I_2, I_3,$ and $I_4$?

(A) In the given figure,the diode in branch $CD$ is reverse-biased because the potential at $C$ is higher than at $D$ relative to the diode orientation. Hence,its resistance is $\infty$,and the current passing through it is $I_3 = 0$. Thus,this branch can be removed from the network.
From the circuit,we have $I_1 = I_2 + I_4$.
The total resistance in branch $AB$ is $R_{AB} = 25\,\Omega + 125\,\Omega = 150\,\Omega$. Similarly,the resistance in branch $EF$ is $R_{EF} = 25\,\Omega + 125\,\Omega = 150\,\Omega$.
Since the branches $AB$ and $EF$ are in parallel and have equal resistances,the current divides equally: $I_2 = I_4 = \frac{I_1}{2}$.
Applying Kirchhoff's voltage law to the loop containing the battery and branches $AB$ and $EF$:
$5\,V - I_1(25\,\Omega) - I_2(150\,\Omega) = 0$
Since $I_1 = I_2 + I_4 = 2I_2$,we substitute $I_1$:
$5 - (2I_2)(25) - 150I_2 = 0$
$5 - 50I_2 - 150I_2 = 0$
$200I_2 = 5$
$I_2 = \frac{5}{200} = 0.025\,A$.
Since $I_2 = I_4$,$I_4 = 0.025\,A$.
$I_1 = I_2 + I_4 = 0.025 + 0.025 = 0.05\,A$.
Therefore,$I_1 = 0.05\,A, I_2 = 0.025\,A, I_3 = 0\,A, I_4 = 0.025\,A$.